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人気のある微分積分 >

integral from 0 to 20 of sqrt(1+4x^2)

解

\int_{0}^{20} 1 + 4 x^{2} d x

解

\frac{1}{2} (201601 + \frac{1}{2} ln (40 + 1601))

+1

十進法表記

401.22052 \dots

解答ステップ

\int_{0}^{20} 1 + 4 x^{2} d x

三角関数による置換を適用する

= \int_{0}^{a r c t a n (40)} \frac{1}{2} sec^{3} (u) d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{a r c t a n (40)} sec^{3} (u) d u

積分換算を適用する

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (40)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (40)} sec (u) d u)

\int_{0}^{a r c t a n (40)} sec (u) d u = ln (40 + 1601)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601))

簡素化

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601))

限度を計算する:

201601

= \frac{1}{2} (201601 + \frac{1}{2} ln (40 + 1601))

グラフ

人気の例

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